Question 4 of 7: Point Defects in Crystalline Solids

Question 4: Point Defects in Crystalline Solids (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

4.1 — Largest Interstitial Void in FCC γ-Fe

Given. FCC γ-Fe, atomic radius $R=0.129$ nm; largest voids at edge-centre $(\tfrac12,0,0)$-type positions.

Find. The void radius $r$.

Approach. In FCC, atoms touch along the face diagonal, $a\sqrt2=4R\Rightarrow a=2\sqrt2R$. The edge-centre void sits half-way along a cell edge (length $a$), touching the two corner atoms; the gap available is $a/2$ minus the host radius.

1. Void radius from cell geometry. $$r=\frac{a}{2}-R=\frac{2\sqrt2R}{2}-R=R(\sqrt2-1)$$ $$r=0.129(\sqrt2-1)=\boxed{0.0534\ \text{nm}}$$

4.2 — Why Point Defects Are Unavoidable, and Their Temperature Dependence

Real crystals always contain a non-zero equilibrium concentration of point defects (vacancies, interstitials) because their presence lowers the crystal's Gibbs free energy, not raises it. Forming a defect costs enthalpy $Q$ (breaking/distorting bonds), but it also increases the crystal's configurational entropy enormously, since a defect can occupy any of $N$ equivalent lattice sites. The free-energy change $\Delta G=\Delta H-T\Delta S$ is minimized at a small but finite defect concentration — a perfect crystal ($\rho=0$) is never the lowest-free-energy state at any $T \gt 0$ K, because the entropy gain from the very first few defects is disproportionately large ($-T\Delta S\to-\infty$ as $\rho\to0$).

Because defect formation is thermally activated, the equilibrium density rises steeply, exponentially, with temperature (appendix form $N_D=N\exp(-Q_d/kT)$):

1. Derive $\rho_1/\rho_2$. With $\rho(T)=N\exp(-Q_v/kT)$ at temperatures $T_1,T_2$: $$\frac{\rho_1}{\rho_2}=\frac{N\exp(-Q_v/kT_1)}{N\exp(-Q_v/kT_2)}=\boxed{\exp\!\left[-\frac{Q_v}{k}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right]}$$

4.3 — Vacancy Concentration in Copper

Given. $Q_v=83{,}680$ J/mol (per-mole form, so $R=8.31$ J/mol·K replaces $k$); $T_1=25\,{}^{\circ}\text{C}=298$ K.

Find. (i) $N_v/N$ at 298 K; (ii) $T_2$ at which $N_v/N$ is 1000× larger.

Approach. Evaluate $N_v/N=\exp(-Q_v/RT)$ at $T_1$; then set the same expression at $T_2$ equal to $1000\times$ the value at $T_1$ and solve for $1/T_2$.

1. Vacancy fraction at 25°C. $$\frac{N_v}{N}=\exp\!\left(-\frac{83{,}680}{(8.31)(298)}\right)=\exp(-33.80)=\boxed{2.11\times10^{-15}}$$
2. Temperature for 1000× the concentration. Setting $\exp(-Q_v/RT_2)=1000\exp(-Q_v/RT_1)$ and taking logs: $$\frac{1}{T_2}=\frac{1}{T_1}-\frac{R\ln(1000)}{Q_v}=3.356\times10^{-3}-6.86\times10^{-4}=2.670\times10^{-3}\ \text{K}^{-1}$$ $$T_2=\boxed{374.6\ \text{K} = 101.6\,{}^{\circ}\text{C}}$$